E4.16 Piping Networks

Example 4.16: A piping network is given below. Find out flow rates through pipes #1, #2, and #3. Pipe #1 (AB) is a horizontal pipe at an elevation of 50 feet. Other end of pipe #2 (BC, point C) has an elevation of 10 feet. Other end of pipe #3 (BD, point D) has an elevation of - 10 feet. Both pipes #2 and #3 have equal pressure drop of 3.7 psi (533 psf). Use the following information:

Roughness of the pipe = 0.00015 ft

Density = 62.37 lb/ft³

Viscosity = 7.53 ´ 10^-4 lb/(ft·s)

Solution: Applying Bernoulli equation between points B and C, frictional loss in pipe #2 can be found to be

Now we can find the value of the group ReÖf by using d₂, and L₂ as

Figure 4.4 can be used to find friction factor as 0.005. This value of friction factor corresponds to Reynolds number of 1.54 ´ 10⁵ (Figure 4.3). Once Reynolds number is known we can find the velocity of the fluid through the pipe #2.

We can do a similar analysis for pipe #3. We can calculate h_f3 to be 68.54 ft·lb_f/lb. Then the group ReÖf for pipe #3 is found to be 7.112 ´ 10³. Figure 4.4 can be used to find friction factor to be 0.006. This value of friction factor corresponds to Reynolds number of 9.08 ´ 10⁴. The velocity of the fluid through pipe #3 can be calculated as 12.54 ft/s. Now knowing the velocities in the pipes #2 and #3 we can calculate the combined flow rate by multiplying velocities in the individual pipes by their cross-sectional areas. This can also result in the velocity of the fluid in pipe #3.

And the combined flow rate is

The Reynolds number can be calculated to be 1.66 ´ 10⁵. Friction factor can be obtained from Figure 4.3 as 0.005. And the frictional losses can be computed to be

Pressure drop in pipe #1 is, therefore,

However, if the pressure is given at point A, C, and D and volumetric flow rates through pipes are required, then pressure at point B is found such that mass is conserved and pressure drop for the estimated flow rates matches with the given pressure. This may involve some iteration.